Question
Expand : $\left(\frac{a}{2}+\frac{b}{3}\right)^2$
✓
Answer
$\left(\frac{a}{2}+\frac{b}{3}\right)^2$
Here $A =\frac{a}{2}$ and $B =\frac{b}{3}$
$
\begin{aligned}
& \left(\frac{ a }{2}+\frac{ b }{3}\right)^2=\left(\frac{ a }{2}\right)^2+2 \times \frac{ a }{2} \times \frac{ b }{3}+\left(\frac{ b }{3}\right)^2 \\
& \text {.... }\left[(A+B)^2=A^2+2 A B+B^2\right] \\
& \therefore \quad\left(\frac{a}{2}+\frac{b}{3}\right)^2=\frac{a^2}{4}+\frac{a b}{3}+\frac{b^2}{9} \\
&
\end{aligned}
$
Here $A =\frac{a}{2}$ and $B =\frac{b}{3}$
$
\begin{aligned}
& \left(\frac{ a }{2}+\frac{ b }{3}\right)^2=\left(\frac{ a }{2}\right)^2+2 \times \frac{ a }{2} \times \frac{ b }{3}+\left(\frac{ b }{3}\right)^2 \\
& \text {.... }\left[(A+B)^2=A^2+2 A B+B^2\right] \\
& \therefore \quad\left(\frac{a}{2}+\frac{b}{3}\right)^2=\frac{a^2}{4}+\frac{a b}{3}+\frac{b^2}{9} \\
&
\end{aligned}
$
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